Name
Sample 2250 Midterm 3 [7:30+10:45, S2009]
Applied Differential Equations 2250
Sample Midterm Exam 3
Exam date: Thursday, 23 April, 2009
Instructions: This in-class exam is 50 minutes. No calculators, notes, tables or
books. No answer check is expected. Details count 75%. The answer counts 25%.
1. (ch5) Complete any combination to make 100%. in order to give you an idea of the range of possible
questions.
(1a) [60%] Find a particular solution yp (x) for
d2 y
d4 y
+
4
= 24x.
dx4
dx2
Answer: y = x3
(1b) [60%] Find the steady-state periodic solution for the forced spring-mass system x00 + 2x0 + 5x =
10 sin(t).
Answer: x = 2 sin t − cos t
(1c) [40%] Given 4x00 (t) + 10x0 (t) + 2wx(t) = 0, which represents a damped spring-mass system with
m = 4, c = 10, k = 2w, determine all values of w such that the equation is under-damped. Do not
solve for x(t)!
Answer: The discriminant is 25 − 8w. This must be negative to produce a complex root, therefore
under-damped happens exactly when w > 25/8.
(1d) [40%] Find by variation of parameters an integral formula for a particular solution xp of the equation
2
x00 + 4x0 + 20x = et ln(t2 + 1). To save time, don’t try to evaluate integrals (it’s impossible).
2
Answer: xp (t) = x1 (t) 0t k1(u)f (u)du+x2 (t) 0t k2(u)f (u)du, f (t) = et ln(t2 +1), x1 (u) = e−2u cos(4u),
x2 (u) = 21 e−2u sin(4u), k1 (u) = −x2 (u)/W (u), k2 (u) = x1 (u)/W (u), W (u) = 2e−4u .
R
R
(1e) [40%] Write the solution of x00 + 9x = 30 sin t, x(0) = x0 (0) = 0, as the sum of two harmonic
oscillations of different natural frequencies. To save time, don’t convert to phase-amplitude
form.
Answer: x(t) = (15 sin t − 5 sin 3t)/4
(1f) [30%] Determine the practical resonance frequency ω for the current equation I 00 + 2I 0 + 10I =
100ω cos(ωt). [original had a typo]
√
√
p
Answer: ω = 1/ LC = 1/ 1/10 = 10.
(1g) [30%] Determine the practical resonance frequency ω for the spring-mass system equation 2x00 +
4x00 + 10x = 100 cos(ωt).
Answer: ω =
p
k/m − c2 /(2m2 ) =
p
5 − 16/8 =
√
3.
2. (ch5) Complete enough parts to make 100%.
(2a) [40%] A non-homogeneous linear differential equation with constant coefficients has right side f (x) =
x2 e−x + x3 (x + 4) + ex cos 2x and characteristic equation of order 8 with roots 0, 0, 0, 1, −1, −1, 2i, −2i,
listed according to multiplicity. Determine the undetermined coefficients shortest trial solution for yp .
To save time, do not evaluate the undetermined coefficients and do not find yp (x)! Undocumented
detail or guessing earns no credit.
Answer: The trial solution is a linear combination of the 9 atoms x2 e−x , x3 e−x , x4 e−x ,
ex cos x, ex sin x.
x3 , x4 , x5 , x6 , x7 ,
(2b) [30%] Find the corrected trial solution in the method of undetermined coefficients for the differential
equation y 00 + y = 3 cos x. To save time, do not evaluate the undetermined coefficients and do not find
yp (x)!
Answer: The trial solution is a linear combination of the atoms x cos x, x sin x.
(2c) [50%] Assume f (x) is a solution of a constant-coefficient linear homogeneous differential equation
whose factored characteristic equation is r2 (r + 1)(r2 + 9) = 0. Find the shortest trial solution in the
method of undetermined coefficients for the differential equation y 000 − y 0 = f (x). To save time, do not
evaluate the undetermined coefficients and do not find yp (x)!
Answer: f (x) must be a linear combination of atoms 1, x, e−x , cos 3x, sin 3x. Equation y 000 − y 0 = 0 has
characteristic equation r3 − r = 0 with roots 0, 1, −1. The trial solution is a linear combination of the atoms
x, x2 , xe−x , cos 3x, sin 3x.
(2d) [50%] Find the shortest trial solution in the method of undetermined coefficients for the differential
equation y 000 − y 0 = x + ex . To save time, do not evaluate the undetermined coefficients and do not
find yp (x)!
Answer: Characteristic equation r3 − r = 0 has roots 0, 1, −1. The trial solution is a linear combination
of the atoms x, x2 , xex .
(2e) [25%] The general solution of a linear homogeneous differential equation with constant coefficients
is
y = c1 ex cos 2x + c2 ex sin 2x + c3 + c4 x + c5 x2 .
Find the factored form of the characteristic polynomial.
Answer: Atom ex cos 2x is constructed from complex root 1 + 2i. Because the conjugate is also a root,
then r − 1 − 2i and r − 1 + 2i are factors and finally the characteristic polynomial is a((r − 1)2 + 4) for some
nonzero constant a.
(2f) [25%] Find six independent solutions of the homogeneous linear constant coefficient differential
equation whose sixth order characteristic equation has roots −1, −1, 0, 0, 2 + i, 2 − i.
Answer:
pendent.
Choose 6 atoms: e−x , xe−x , 1, x, e2x cos x, e2x sin x. Cite a theorem: Distinct atoms are inde-
(2g) [25%] Let f (x) = x3 + x sin 2x. Find a constant-coefficient linear homogeneous differential equation
which has f (x) as a solution.
Answer: The atom x3 is constructed from roots 0, 0, 0, 0 corresponding to factor r4 . The atom x sin 2x is
constructed from roots ±2i, ±2i corresponding to factor (r2 +4)2 . The characteristic polynomial r4 (r2 +4)2 =
r8 + 8r6 + 16r4 belongs to DE y (8) + 8y (6) + 16y (4) = 0.
(2h) [20%] Let f (x) = 4ex − cosh x + ex cos2 2x. Find the characteristic polynomial of a constantcoefficient linear homogeneous differential equation which has f (x) as a solution.
Answer: Because cosh x = (ex + e−x )/2, then roots 1, −1 are used to produce atoms for the first two
terms 4ex and − cosh x. Because cos2 2x = (1 + cos 4x)/2 [identity cos 2θ = 2 cos2 θ − 1], then roots 1 ± 4i
and 0 are used to produce the third term ex cos2 2x. Total roots: 1, −1, 0, 1 ± 4i with product of the factors
(r−1)(r+1)(r−0)((r−1)2 +16) = r5 −2r4 +16r3 +2r2 −17r. The DE: y (5) −2y (4) +16y 000 +2y 00 −17y 0 = 0.
3. (ch10) Complete enough parts to make 100%. It is assumed that you have memorized the basic 4-item
Laplace integral table and know the 6 basic rules for Laplace integrals. No other tables or theory
are required to solve the problems below. If you don’t know a table entry, then leave the expression
unevaluated for partial credit.
(3a) [50%] Apply Laplace’s method to solve the system for x(t). Don’t waste time solving for y(t)!
x0 = 3y,
y 0 = 2x − y,
x(0) = 0, y(0) = 1.
Answer: x(t) = − 35 e−3t + 53 e2t , y(t) = !53 e−3t + 25 e2t . Obtained by the Laplace resolvent method to save
x(t)
time: (A − sI)L(u) = u(0), u =
.
y(t)
(3b) [25%] Find f (t) by partial fraction methods, given
L(f (t)) =
3s + 4
.
s(s + 4)
Answer: L(f (t)) = 1/s + 2/(s + 4) = L(1 + 2e−4t )
(3c) [25%] Solve for f (t), given
d
L(f (t)) = L te2t s→(s+1)
ds
Answer: −tf (t) = e−t te2t or f (t) = −et .
(3d) [25%] Solve for f (t), given
L(f (t)) =
Answer: L(f (t)) =
s+1
s
2 1
s
1
s
=
s→(s+1)
+
2
s2
s+2
s+1
+
2
1
s+1
1
s3 s→(s+1)
= L((1 + 2t + t2 /2)e−t ).
4. (ch10) Complete enough parts to make 100%.
(4a) [25%] Fill in the blank spaces in the Laplace table:
f (t)
t3
L(f (t))
6
s4
Answers: Left to right:
t2 e−t
et cos t
1
2s + 2
1 −t
2e ,
e−t cos 3t,
s2
s+1
+ 2s + 10
s−1
, 2 .
(s−1)2 +1 (s+1)3
(4b) [50%] Solve by Laplace’s method for the solution x(t):
x00 (t) + x0 (t) = 9e−t ,
x(0) = x0 (0) = 0.
Answer: x(t) = −(9 + 9t)e−t + 9.
(4c) [25%] Solve for x(t), given
L(x(t)) =
d L(e2t sin 2t) + L(t sin t)|s→(s+2) .
ds
Answer: x(t) = −te2t sin 2t + te−2t sin t.
(4d) [25%] Solve for x(t), given
L(x(t)) =
Answer: L(x(t)) =
s+1 s2 s→(s+1)
+
1
5s
+
2t
(4e) [25%] Find L(f (t)), given f (t) = e
4
5(s+4)
s+2
1+s
+ 2
2
(s + 1)
s + 5s
= L (1 + t)e−t +
sin(t)
.
t
1
5
+ 54 e−4t .
d
Answer: Let g(t) = (1/t) sin t. Then ds
L(g(t)) = L(−tg(t)) = L(− sin t) = −1/(s2 + 1). Solve
dG(s)
π
−1
π
ds = s2 +1 to get G(s) = 2 − arctan s. Then L(f (t)) = L(g(t))|s→(s−2) = 2 − arctan(s − 2).
(4f) [30%] Apply Laplace’s method to find a formula for L(x(t)). Do not solve for x(t)! Document
steps by reference to tables and rules.
d4 x
d2 x
+
4
= et (5t + 4et + 3 sin 3t),
dt4
dt2
x(0) = x0 (0) = x00 (0) = 0,
x000 (0) = −1.
t
t
4
2
Answer:
L(x(t))
= p/q, p = 1 + L(e (5t + 4e + 3 sin 3t)), q = s + 4s . Expanding, p = 1 +
5
5
4
4
9
= 1 + (s−1)
+ s−1
+ s29+9 2 + s−2 + (s−1)2 +9 .
s2
s→(s−1)
(4g) [25%] Find L(f (t)), given f (t) = u(t − π)
sin(t)
, where u is the unit step function [original had a
t−π
typo].
Answer:
Use
theorem L(u(t − a)g(t)) = e−as L(g(t + a)). Then L(f (t)) =
the second
shifting
sin(t+π)
t
t
−1
e−πs L t−π+π = e−πs L − sin
. Let h(t) = − sin
t
t , then L(th(t)) = s2 +1 implies a first order DE
− dH
ds =
−1
,
s2 +1
which can be solved to get H(s) = arctan(s) − π2 . Then L(f (t)) = e−πs (arctan(s) − π/2).
(4h) [40%] Find f (t) by partial fraction methods, given
L(f (t)) =
Answer: L(f (t)) =
−1
s−1
+
−3
s+3
+
4
(s+1)2
+
8s2 − 24
.
(s − 1)(s + 3)(s + 1)2
4
s+1
= L −et − 3e−3t + 4te−t + 4e−t .
(4i) [30%] Apply Laplace’s method to find a formula for L(x(t)). To save time, do not solve for x(t)!
Document steps by reference to tables and rules.
x(4) − x(2) = 3t2 + 4e−2t + 5et sin 2t,
x(0) = x0 (0) = x00 (0) = 0,
x000 (0) = −1.
Answer: L(x(t)) = p/q, p = −1 + L(RHS), q = s4 − s2 . Finally p = −1 + L(3t2 + 4e−2t + 5et sin 2t) =
4
10
+ (s−1)
−1 + s63 + s+2
2 +4 .
5. (ch6) Complete all of the items below.




(5a) [30%] Find the eigenvalues of the matrix A = 
eigenvectors!
Answer: −1, 5, 3 ± 2i
−2
−1
0
0

7
1 12
6 −3 15 

. To save time, do not find
0
3 2 
0 −2 3


5 1 −1


1 , assume there exists an invertible matrix P and a diagonal matrix
(5b) [40%] Given A =  0 3
0 1
3
D such that AP = P D. Circle all possible columns of P from the list below.






1


 1 .
0
0


 0 ,
1
2


 −3  ,
3
Answer: Matrix P must contain eigenvectors of P corresponding to eigenvalues 2, 4, 5. For each given
vector v, multiply Av and see if it is λv for some λ. The first works fir λ = 2 and the others do not work.
1 −1
−2
2
(5c) [30%] Find all eigenpairs for the matrix A =
Answer:
0,
1
1
!!
,
3,
!
. Then display Fourier’s model for A.
!!
1
−2
.
(5d) [50%] Find the remaining eigenpairs of


6 2 −2


1 
E= 0 5
0 1
5
provided we already know one eigenpair



0

 
6,  1  .
1
 

 

0
1
2

 
 

Answer: Eigenvalues are 4, 6, 6 with corresponding eigenvectors  −1 ,  0 ,  1 .
1
0
1
(5e) [25%] Suppose a 2 × 2 matrix A has eigenpairs
3,
3
1
!!
, −2,
1
−2
!!
. Display an invertible
matrix P and a diagonal matrix D such that AP = P D.
3
1
1 −2
Answer: Define P =
!
3
0
0 −2
,D=
!
. Then AP = P D.
(5f) [25%] Assume the vector general solution x(t) of the linear differential system x0 = Ax is given by






0
3
−1




2t 
2t 
x(t) = c1  1  + c2 e  2  + c3 e  0  .
1
0
1
Display Fourier’s model for the 3 × 3 matrix A.













−1
0
3
−1
0
3
 











Answer: A c1  1  + c2  2  + c3  0  = 0c1  1  + 2c2  2  + 2c3  0 .
1
0
1
1
0
1


−2
−1
0
0



(5g) [30%] Find the eigenvalues of the matrix A = 
7
1 27
6 −3 62 

. To save time, do not find
0
3 2 
0 −1 0
eigenvectors!
Answer: Expand by cofactors along column 1. The eigenvalues are −1, 1, 2, 5.
(5h) [30%] Assume A is 2 × 2 and Fourier’s model holds:
1
1
A c1
!
1
−1
+ c2
!!
1
−1
= 2c2
!
.
Find A.
Answer: AP = P D implies A =

P DP −1
1
1
1 −1
=
!
0 0
0 2
!
.5
.5
.5 −.5
!
1 −1
−1
1
=
!
.

3 0 −1


0 . Circle the possible eigenvectors of A in the list below.
(5i) [40%] Let A =  0 3
0 0
3



−4


 2 ,
0

1


 0 ,
0


0


 0 .
1
Answer:
Fourier’s
 model does not hold [A is not diagonalizable] because there are only two eigenvectors



0
1




 0  and  1  for eigenvalue λ = 3. The first is a linear combination of these eigenvectors, hence itself
0
0
an eigenvector. The second is one already reported, The third is not an eigenvector. The problem should be
solved by testing the equation Av = 3v for each of the 3 vectors v in the list, not by doing the eigenanalysis
of A.
(5j) [40%] Consider the 3 × 3 matrix


4 2 −2


1 .
E= 0 3
0 1
3
Show that matrix E has a Fourier model: [original had a typo]













1
0
2
1
0
2
 











E c1  0  + c2  1  + c3  −1  = 4c1  0  + 4c2  1  + 2c3  −1  .
0
1
1
0
1
1
Answer: Do the eigenanalysis of A. Alternate: verify that the eigenpairs extracted from Fourier’s model
actually work, which involves 3 matrix multiplies.


2


3
(5k) [40%] Define E as in the previous problem. Find E  0  without using matrix multiply.
2

2

Answer: Vector v ≡  0
2

16

3
3
3
E v = 4 v2 + 2 v3 =  56
72
(5l) [30%] Let P =


 = v2 + v3 , where v1 , v2 , v3 are the eigenvectors in Fourier’s model. Then
3
1
1 −1


.
!
,D=
3
0
0 −2
!
and define A by AP = P D. Display the eigenpairs of
A.
Answer:
3,
3
1
!!
,
−2,
1
−1
!!
(5m) [30%] Assume the vector general solution x(t) of the linear differential system x0 = M x is given
by
!
!
2
−1
−t
2t
x(t) = c1 e
+ c2 e
.
1
1
Display an invertible matrix P and a diagonal matrix D such that M P = P D.
Answer: !!
The eigenvalues
come from the coefficients
−1 and 2. The eigenpairs are
!!
! in the exponentials,
!
2
−1
2 −1
−1 0
−1,
, 2,
. Then P =
,D=
.
1
1
1
1
0 2